Calculate the standard cell potentials of galvanic cell in whiCHM the following reactions take place `:`
`a. Cr(s) +3Cd^(2+)(aq) rarr 2Cr^(3+)(aq)+3Cd`
`b. Fe^(2+)(aq)+Ag^(o+)(aq)rarr Fe^(3+)(aq)+Ag(s)`
Calculate the `Delta_(r)G^(c-)` and equilibrium constant of the reactions .

Explanation

To solve the problem, we will calculate the standard cell potentials (E°) for the given reactions, followed by the Gibbs free energy change (ΔG°) and the equilibrium constant (K) for each reaction.

### Step 1: Identify the half-reactions and their standard reduction potentials.

**a. Reaction:**
\[ \text{Cr(s)} + 3\text{Cd}^{2+}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 3\text{Cd(s)} \]

**Half-reactions:**
1. Oxidation:
\[ \text{Cr(s)} \rightarrow \text{Cr}^{3+}(aq) + 3e^{-} \]
(Standard reduction potential, E° = -0.74 V)

2. Reduction:
\[ \text{Cd}^{2+}(aq) + 2e^{-} \rightarrow \text{Cd(s)} \]
(Standard reduction potential, E° = +0.40 V)

**b. Reaction:**
\[ \text{Fe}^{2+}(aq) + \text{Ag}^{+}(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{Ag(s)} \]

**Half-reactions:**
1. Oxidation:
\[ \text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^{-} \]
(Standard reduction potential, E° = -0.77 V)

2. Reduction:
\[ \text{Ag}^{+}(aq) + e^{-} \rightarrow \text{Ag(s)} \]
(Standard reduction potential, E° = +0.80 V)

### Step 2: Calculate the standard cell potential (E°cell).

**a. For the first reaction:**
\[ E°_{cell} = E°_{cathode} - E°_{anode} \]
\[ E°_{cell} = E°_{Cd/Cd^{2+}} - E°_{Cr/Cr^{3+}} \]
\[ E°_{cell} = 0.40 \, \text{V} - (-0.74 \, \text{V}) \]
\[ E°_{cell} = 0.40 + 0.74 = 1.14 \, \text{V} \]

**b. For the second reaction:**
\[ E°_{cell} = E°_{Ag/Ag^{+}} - E°_{Fe/Fe^{3+}} \]
\[ E°_{cell} = 0.80 \, \text{V} - (-0.77 \, \text{V}) \]
\[ E°_{cell} = 0.80 + 0.77 = 1.57 \, \text{V} \]

### Step 3: Calculate ΔG° for each reaction.

The relationship between ΔG° and E°cell is given by:
\[ \Delta G° = -nFE°_{cell} \]
where:
- n = number of moles of electrons transferred in the balanced equation
- F = Faraday's constant (approximately 96485 C/mol)

**a. For the first reaction:**
- n = 6 (3 moles of Cd²⁺ each gaining 2 electrons)
\[ \Delta G° = -6 \times 96485 \, \text{C/mol} \times 1.14 \, \text{V} \]
\[ \Delta G° = -6 \times 96485 \times 1.14 \approx -660,000 \, \text{J} \]
\[ \Delta G° \approx -660 \, \text{kJ} \]

**b. For the second reaction:**
- n = 1 (1 mole of Fe²⁺ losing 1 electron)
\[ \Delta G° = -1 \times 96485 \, \text{C/mol} \times 1.57 \, \text{V} \]
\[ \Delta G° = -96485 \times 1.57 \approx -151,000 \, \text{J} \]
\[ \Delta G° \approx -151 \, \text{kJ} \]

### Step 4: Calculate the equilibrium constant (K) for each reaction.

The relationship between ΔG° and K is given by:
\[ \Delta G° = -RT \ln K \]
where:
- R = 8.314 J/(mol·K)
- T = temperature in Kelvin (assume 298 K)

**a. For the first reaction:**
\[ -660,000 = -8.314 \times 298 \ln K \]
\[ \ln K = \frac{660000}{8.314 \times 298} \approx 265.7 \]
\[ K = e^{265.7} \]

**b. For the second reaction:**
\[ -151,000 = -8.314 \times 298 \ln K \]
\[ \ln K = \frac{151000}{8.314 \times 298} \approx 60.7 \]
\[ K = e^{60.7} \]

### Summary of Results:
1. **First Reaction:**
- E°cell = 1.14 V
- ΔG° ≈ -660 kJ
- K ≈ e^{265.7}

2. **Second Reaction:**
- E°cell = 1.57 V
- ΔG° ≈ -151 kJ
- K ≈ e^{60.7}